I need to do some iterative summations. Here is a minimum working example:

data = Table[RandomReal[], x, 1, 1000000];
(* Method 1 *)
Timing[Total[Table[Total[ Table[data[[i]], i, j, 10 + j]], j, 1, Length[data] - 5*10]]]

(* Method 2, with constant index *)
m = 10;
Timing[Total[Table[Total[ Table[data[[i]], i, j, m + j]], j, 1, Length[data] - 5*m]]]

And here are the outputs:

0.5625, 5.49936*10^6

9.28125, 5.49936*10^6

For some reason, using m=10 makes it much slower. I will need to do a bunch of m's, so this is the bottom of a larger nest.

What is a faster way to do this?

Late Edit:

Bonus question: How to optimize this one as well:

Timing[Total[Table[ (Total[ Table[data[[i]], i, j, m + j]])^2 , j, 1, Length[data] - 5*m]]]

The problem lies mostly in the inner Table:

Timing[Total[Table[Total[Table[data[[i]], i, j, 10 + j]], j, 1,Length[data] - 5*10]]]

m = 10;
Timing[Total[Table[Total[Table[data[[i]], i, j, m + j]], j, 1, Length[data] - 5*10]]]

0.366407, 5.50276*10^6

8.01738, 5.50276*10^6

I think the reason is this:
Because the global variable m could theoretically change its value during the computions, the body of the outer table cannot be compiled (without calls to MainEvaluate). At least, the JIT compiler does not analyze the body of the outer loop thoroughly enough to decide that m won't change.

You can help the JIT compiler by using With:

With[m = 10,
 Timing[Total[Table[Total[Table[data[[i]], i, j, m + j]], j, 1,Length[data] - 5*m]]]
 ]

0.369601, 5.5049*10^6

Addendum:

By focusing on the post's title, I have completely overlooked the question on how to make it faster. Here is my proposal (c) vs. the OP's one (a) and Carl's (b):

a = With[m = 10,
 Total[
 Table[Total[Table[data[[i]], i, j, m + j]], j, 1, 
 Length[data] - 5*m]]
 ]; // RepeatedTiming // First
b = Total@ListCorrelate[ConstantArray[1., m + 1], 
 data[[;; -50 + m - 1]]]; // RepeatedTiming // First
c = Plus[
 Range[1., m].data[[1 ;; m]],
 (m + 1) Total[data[[m + 1 ;; -5*m - 1]]],
 Range[N@m, 1., -1].data[[-5 m ;; -4 m - 1]]
 ]; // RepeatedTiming // First

a == b == c

0.28

0.017

0.0018

True

You can use ListCorrelate:

m=10;
Total @ ListCorrelate[ConstantArray[1,m+1], data[[;;-4 m-1]]] //AbsoluteTiming

0.017725, 5.50044*10^6

Bonus question

For the bonus question:

data = RandomReal[1, 10^5];

Your version:

With[m = 10,
 Total[Table[(Total[Table[data[[i]],i,j,m+j]])^2,j,1,Length[data]-5*m]]
] //AbsoluteTiming

0.448739, 3.11778*10^7

Using ListCorrelate again:

m = 10;
#.#& @ ListCorrelate[ConstantArray[1, m+1], data[[ ;; -4 m - 1]]] //AbsoluteTiming

0.018401, 3.11778*10^7